Completing the square uses half the linear coefficient because squaring a binomial creates two matching linear terms. In (x + t)², those terms are tx and tx, giving 2tx. To match bx, set 2t = b, so t = b/2. The missing constant is then t² = (b/2)².
The half tells you what belongs inside the parentheses. The square tells you what constant to add outside them. Mixing those two jobs is why adding 3 to x² + 6x doesn’t complete the square. You need 9, because the matching binomial is x + 3.
This lesson explains where the rule comes from, how a square picture supports it, and how to keep your original expression or equation equal. If coefficients and factoring still feel unfamiliar, the algebra lessons and tools provide a useful starting point.
Why Does Completing The Square Use Half The Linear Coefficient?
The coefficient gets halved because the middle term of a squared binomial comes from two products. Expand (x + t)(x + t), and you get x² + tx + tx + t². Each copy contributes tx, so the total linear term is 2tx. Matching bx requires 2t = b.
Here, t stands for the number we haven’t found yet. It could be positive, negative, or a fraction. Using a separate letter helps us distinguish that number from b, which already names the coefficient of x in the original expression.
(x + t)² = x² + 2tx + t²
2t = b, so t = b/2
x² + bx + (b/2)² = (x + b/2)²
OpenStax’s completing the square lesson presents the binomial square pattern and the rule of adding the square of half the linear coefficient. The expansion above explains why that rule works: matching the middle term determines the number inside the binomial.
Why Not Square The Whole Coefficient?
Using the whole coefficient would double the middle term you wanted. For x² + 6x, choosing t = 6 produces (x + 6)² = x² + 12x + 36. The 12x doesn’t match 6x. Choosing t = 3 instead produces x² + 6x + 9, which matches both variable terms.
A quick check is to expand your proposed square before doing anything else. The coefficient of x should stay exactly the same. If it changed, the number inside the parentheses is wrong. This check works even when your arithmetic involves fractions or negative signs.
The original constant doesn’t choose the half. In x² + 6x + 2 and x² + 6x + 20, the number inside the square is still 3. Only the leftover constant changes. The first expression becomes (x + 3)² − 7; the second becomes (x + 3)² + 11.
How Does The Missing Corner Explain The Rule?
A square picture explains the half by dividing the linear area into two equal strips. For x² + 6x, split the 6x into 3x and 3x. Place one strip beside an x-by-x square and the other below it. A 3-by-3 corner completes the larger square.
The two strips explain why we divide by 2 before we square. Each strip has width 3, so the missing corner has side length 3 on both edges. Its area is 3 × 3 = 9. The completed figure has side length x + 3.
Why wouldn’t one strip of width 6 work? It would form a rectangle with sides x and x + 6. Splitting the strip gives both sides the same added length. Completing the corner then turns that arrangement into a square, which we can describe with one squared binomial.
The picture supports the algebra, but it has limits. Physical lengths and areas here are positive. A negative linear coefficient doesn’t mean a tile has negative area. Use the multiplication identity for negative coefficients; it applies to all real values without needing physical strips.
What Happens With Negative Or Fractional Coefficients?
The same rule works with negative and fractional coefficients: keep the sign when halving b, then square the result. For x² − 10x, half of −10 is −5. Its square is 25, so x² − 10x + 25 = (x − 5)². The added constant is positive.
Expanding confirms the sign: (x − 5)(x − 5) gives x² − 5x − 5x + 25. Both linear products are negative, but multiplying −5 by −5 makes the constant positive. The sign inside the binomial follows b, while its square supplies the missing constant.
Odd Coefficients Give Exact Fractions
For x² + 3x, half of 3 is 3/2. Squaring gives (3/2)² = 9/4. Therefore, x² + 3x + 9/4 = (x + 3/2)². A fraction isn’t a problem to remove. It’s the exact number needed to keep the two linear products equal to 3x.
Keep fractions until the last step, especially when solving an equation. Rounding 3/2 early might seem harmless here, but other fractions don’t have short decimal forms. Exact fractions also make expansion easier to check: 2 × (3/2)x = 3x, so the linear term matches immediately.
| Starting terms | Half of b | Constant to add | Completed square |
|---|---|---|---|
| x² + 6x | 3 | 9 | (x + 3)² |
| x² − 10x | −5 | 25 | (x − 5)² |
| x² + 3x | 3/2 | 9/4 | (x + 3/2)² |
Keeping Expressions And Equations Equal
Adding the completing term changes a value unless you balance it. In an equation, add the same number to both sides. In an expression, add and subtract that number within the expression. These are two ways to preserve equality, depending on whether you’re solving or rewriting.
Solving An Equation
Start with x² + 6x = 7. The linear coefficient is 6, so half is 3 and its square is 9. Add 9 to both sides. The left becomes a perfect square, while the right becomes 16. Both sides changed by the same amount.
x² + 6x + 9 = 7 + 9
(x + 3)² = 16
x + 3 = 4 or x + 3 = −4
x = 1 or x = −7
Both answers pass the original equation. For x = 1, we get 1² + 6(1) = 1 + 6 = 7. For x = −7, we get (−7)² + 6(−7) = 49 − 42 = 7. Checking both catches sign errors after the square root step.
The symbol √16 names the principal square root, which is 4. However, solving a square equation asks for every number whose square is 16. Both 4 and −4 qualify. The principal square root explanation shows why these two questions have different answers.
Rewriting An Expression
For the expression x² + 6x, there’s no other side to balance. Write x² + 6x + 9 − 9 instead. The added pair totals zero, preserving the value. Then group the first three terms: x² + 6x = (x + 3)² − 9.
Don’t write x² + 6x = (x + 3)² by itself. At x = 0, the left equals 0 and the right equals 9. The outside −9 fixes that difference for every x, not just for a particular test value.
What If The Coefficient Of x² Is Not 1?
First make the coefficient of x² equal to 1 inside the part you complete. In 2x² + 8x + 3, factor 2 from the variable terms to get 2(x² + 4x) + 3. The coefficient to halve is now 4, so the completing constant inside is 4.
2x² + 8x + 3 = 2(x² + 4x + 4 − 4) + 3
= 2[(x + 2)² − 4] + 3
= 2(x + 2)² − 5
The outside 2 multiplies the subtracted 4 too. That makes the correction −8, which combines with +3 to give −5. Subtracting only 4 outside the parentheses would change the original expression. Always track the multiplier when you move the balancing constant outside.
If y = 2(x + 2)² − 5, the squared part reaches zero when x = −2. Every other real x makes it positive. Therefore, the graph’s lowest point, called its vertex, is (−2, −5). Completing the square reveals both coordinates directly.
For an equation with a leading coefficient other than 1, you can divide both sides by that coefficient first. Divide every term, including the constant and the right side. The coefficient you halve is the new coefficient of x after division.
In general, factoring a from ax² + bx gives a[x² + (b/a)x], provided a is not zero. Inside the brackets, half the linear coefficient is b/(2a). This explains why halving the original b too soon fails: the coefficient of x must be compared with a square whose x² coefficient is 1.
Common Questions About Completing The Square
The most useful checks are simple: match the linear term, preserve the value, and keep both possible roots. You can check the rewritten form with the completing the square calculator, then expand it yourself to see why each coefficient matches.
What If b Is Zero?
No extra term is needed. Half of zero is zero, and its square is zero. The expression x² is already a perfect square. For x² + 7, the square part remains x² and the constant stays outside. There’s no linear term to rebuild.
Can The Completing Constant Be Negative?
For real b, the completing constant (b/2)² cannot be negative. A negative b changes the sign inside the binomial, such as x − 5, but squaring −5 still gives +25. A negative outside constant may remain after balancing the original expression.
What If The Completed Square Equals A Negative Number?
There are no real solutions. For example, (x + 3)² = −4 cannot hold for any real x, because a real number squared is never negative. Complex numbers extend what we can solve, but they are outside this lesson’s scope.
How Does This Connect To The Quadratic Formula?
The quadratic formula comes from completing the square in a general quadratic equation. The same middle-term matching explains the half there too. Once this step makes sense, the separate quadratic formula derivation shows how the rest of that proof follows.
When practicing, write the half and its square on separate lines. Then decide how to balance the addition. Finally, expand your completed square or substitute your solutions into the original equation. Those checks turn a memorized rule into a method you can explain.